How the time_point created with different duration(std::chrono::milliseconds and std::chrono::nanoseconds) is so different

Question

I have created std::chrono::milliseconds ms and std::chrono::nanoseconds ns from std::chrono::system_clock::now().time_since_epoch(). From that duration I created timepoints and convert it to time_t using system_clock::to_time_t and print it using ctime function. But the time printed is not same. As I understand the time_point have duration and duration have rep and period (ratio). So time_point must have same value up to millisecond precision in both time_points. Why the output is different?

Here is my code

#include <ctime> #include <ratio> #include <chrono> #include <iostream> using namespace std::chrono; int main () { std::chrono::milliseconds ms = std::chrono::duration_cast < std::chrono::milliseconds > (std::chrono::system_clock::now().time_since_epoch()); std::chrono::nanoseconds ns = std::chrono::duration_cast< std::chrono::nanoseconds > (std::chrono::system_clock::now().time_since_epoch()); std::chrono::duration<unsigned int,std::ratio<1,1000>> today_day (ms.count()); std::chrono::duration<system_clock::duration::rep,system_clock::duration::period> same_day(ns.count()); system_clock::time_point abc(today_day); system_clock::time_point abc1(same_day); std::time_t tt; tt = system_clock::to_time_t ( abc ); std::cout << "today is: " << ctime(&tt); tt = system_clock::to_time_t ( abc1 ); std::cout << "today is: " << ctime(&tt); return 0; }

Howard Hinnant · Accepted Answer · 2018-07-26 13:04:16Z

This line:

std::chrono::duration<unsigned int,std::ratio<1,1000>> today_day (ms.count());

is overflowing. The number of milliseconds since 1970 is on the order of 1.5 trillion. But unsigned int (on your platform) overflows at about 4 billion.

Also, depending on your platform, this line:

std::chrono::duration<system_clock::duration::rep,system_clock::duration::period> same_day(ns.count());

may introduce a conversion error. If you are using gcc, system_clock::duration is nanoseconds, and there will be no error.

However, if you're using llvm's libc++, system_clock::duration is microseconds and you will be silently multiplying your duration by 1000.

And if you are using Visual Studio, system_clock::duration is 100 nanoseconds and you will be silently multiplying your duration by 100.

Here is a video tutorial for <chrono> which may help, and contains warnings about the use of .count() and .time_since_epoch().

Sir i hava a question is the time_since_epoch is the duration between reference point(1970-01-01T00:00:00Z) and current UTC time or it is the duration between reference point(1970-01-01T00:00:00Z) and current local time.I guess it is the first one.
Officially the system_clock::time_since_epoch() is unspecified. However all implementations are measuring the current UTC time since 1970-01-01T00:00:00Z, excluding leap seconds. That is, in this measure, every day is exactly 86400 seconds. This is known as Unix Time. The current C++20 draft specification makes this definition official.
Why hasn't this tutorial be linked in all the other chrono Q&A threads? Thanks a lot! And now I know, why I see your name on all the (helpful) answers I've been studying today... Wish I could +1 it more than once.

Maxim Egorushkin · Accepted Answer · 2018-07-26 12:08:02Z

The conversions you do manually do not look right.

You should use duration_cast for conversions because they are type-safe:

auto today_day = duration_cast<duration<unsigned, std::ratio<86400>>>(ms); auto same_day = duration_cast<system_clock::duration>(ns);

Outputs:

today is: Thu Jul 26 01:00:00 2018 today is: Thu Jul 26 13:01:08 2018

Caleth · Accepted Answer · 2018-07-26 12:08:03Z

Because you throw away the duration info, and then interpret an integer value as a different duration type

std::chrono::duration<unsigned int,std::ratio<1,1000>> today_day (ms.count());

milliseconds -> dimensionless -> 1 / 1000 seconds (i.e. milliseconds)

std::chrono::duration<system_clock::duration::rep,system_clock::duration::period> same_day(ns.count());

nanoseconds -> dimensionless -> system clocks

You should instead just duration_cast again

#include <ctime> #include <ratio> #include <chrono> #include <iostream> using namespace std::chrono; int main () { milliseconds ms = duration_cast<milliseconds>(system_clock::now().time_since_epoch()); nanoseconds ns = duration_cast<nanoseconds>(system_clock::now().time_since_epoch()); system_clock::time_point abc(duration_cast<system_clock::duration>(ms)); system_clock::time_point abc1(duration_cast<system_clock::duration>(ns)); std::time_t tt; tt = system_clock::to_time_t ( abc ); std::cout << "today is: " << ctime(&tt); tt = system_clock::to_time_t ( abc1 ); std::cout << "today is: " << ctime(&tt); return 0; }

Thanks for your response.I think for my code it will be better it will fail at compile time if i did not pass system_clock::duration to system_clock::time_point at constructor.
Yes i got it. But i think it will be better if the code i posted failed at compile time if as i did not pass system_clock::duration to system_clock::time_point at constructor. It can check the rep and period of passed duration is matched with system_clock::duration::rep and system_clock::duration::period if not failed at compile time.

Collectives™ on Stack Overflow

How the time_point created with different duration(std::chrono::milliseconds and std::chrono::nanoseconds) is so different

3 Answers 3

5 Comments

Comments

3 Comments

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

5 Comments

Comments

3 Comments

Related